# Express a irreducible polynomial $f(X)$ in the field of characteristic $p>0$ $F[X]$ as $g(X^{p^m})$

$$F[X]$$ is of the form $$a_0+a_1X+...+a_nX^n$$ where $$a_0,a_1,...a_n\in F\text{ with characteristic p>0}.$$ Express $$f\in F[X]\ as\ g(X^{p^m})$$, where the nonnegative integer m is a large as possible and $$f$$ is irreducible. Show that g is irreducible and separable.

This is a problem in section 3.4 of Basic Algebra by Robert Ash.

The question I had about is how to write $$f(X)$$ in form of $$g(X^{p^m})$$. I have tried this using the property $$(a+b)^p=a^p+b^p$$, since $$F$$ has characteristic p and the Frobenius Automorphism indicate if $$\alpha \in F, \alpha = \beta ^p,$$ for some $$\beta \in F$$. $$g(X^{p^m})$$ can be therefore written as $$(b_0+b_1X+...+b_nX^n)^{p^m}.$$ This is writing $$g(X^{p^m})$$ in in form of $$f(X)$$, for the other way around, is it just taking the $$p^m$$ root of the above expression: $$f(X)=\sqrt[p^m]{g(X^{p^m})}$$

• As you state this it is not true: for $p=2$, $f=X^4$, $g=X^2$ we have $f(X)=g(X^2)$ but $g$ is not irreducible. Perhaps you can type the question exactly it is in Ash? Feb 13, 2021 at 8:31
• Thank you for reminding, I will change the question. Feb 13, 2021 at 8:34
• But it is still false. Take $f=X(X+1)$, then we must have $g=f$ (with $m=0$) and $g$ is not irreducible. There must be more to the question than this. Is $f$ just possibly meant to be the minimal polynomial of some element in an extension field? Feb 13, 2021 at 9:21
• I add the condition $f$ is irreducible Feb 13, 2021 at 19:37

The fact that $$g$$ is irreducible follows from the fact that $$f$$ is; i.e. if $$g = h_1 h_2$$ then $$f = h_1(x^{p^m})h_2(x^{p^m})$$.
Once we show existence, this proves that $$g$$ is separable too: an irreducible polynomial is separable iff $$g' \neq 0$$, and in characteristic $$p$$ the fact that $$g' = 0$$ in turn means that $$g = \widetilde{g}(x^p)$$ for some $$\widetilde g$$. We cannot have the latter, otherwise it would contradict the maximality of $$m$$.
$$S = \{m \in \mathbb N_0 : f = g(x^{p^m}), \text{ for some g}\}.$$
The set $$S$$ is non empty, because $$0 \in S$$, and it is bounded above by degree considerations. Hence there exists $$m = \max S$$ with an associated $$g$$, which is irreducible and separable by the previous remarks.